IMAT Chemistry · Chemical Bonding

VSEPR Shapes, Bond Angles and Polar Molecules

Electron pairs around a central atom push as far apart as possible, which fixes the shape; the shape then decides whether the bond dipoles cancel.

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Why this matters

This is the most asked page of the chapter. Shapes and bond angles came up in 2016, 2017, 2018, 2021 and 2022, and "which molecule has a permanent dipole" in 2014, 2018, 2019, 2020 and the 2023 ministry paper.

Concept 1 of 3: Counting electron domains on the central atom (VSEPR)

Every group of electrons around the central atom, whether a bond or a lone pair, is a cloud of negative charge. The clouds repel each other and spread out as far as they can. So the number of clouds fixes the arrangement, and you only need to count them.

Definition

VSEPR (valence shell electron pair repulsion): the electron domains around a central atom arrange themselves as far apart as possible.

A domain is a single bond, a double bond, a triple bond or a lone pair: a multiple bond counts as one domain.
Number of domains = atoms bonded to the central atom + lone pairs on it.
The shortcut formula below gives the number of domains directly. Then lone pairs = domains minus atoms bonded.
The shape is named from the positions of the atoms only; lone pairs shape the molecule but are not part of its name.

Domains on the central atom

D = \frac{V + m - q}{2}

$V$ valence electrons of the central atom
$m$ number of H or halogen atoms bonded to it (O and S atoms add nothing)
$q$ charge of the ion, with its sign (so a 2− ion adds 2)
$D$ number of electron domains; round a half up (odd-electron molecules)

Worked example

Find the number of lone pairs on the central atom and the shape of (a) nitrogen trichloride,

\mathrm{NCl_3}

, and (b) the carbonate ion,

\mathrm{CO_3^{2-}}

Practice this conceptself-check · 4 quick reps

Forgetting the charge changes the shape

\mathrm{SO_3}

is trigonal planar but

\mathrm{SO_3^{2-}}

is trigonal pyramidal: the two extra electrons make a lone pair. Always add the negative charge (or subtract the positive charge) before counting domains.

Concept 2 of 3: The VSEPR shapes and their bond angles

Two domains go to opposite sides, three to the corners of a triangle, four to the corners of a tetrahedron. Lone pairs sit closer to the central atom than bonding pairs and push harder, so each lone pair squeezes the bond angles a little below the ideal value.

Definition

Repulsion order: lone pair to lone pair > lone pair to bond pair > bond pair to bond pair.

Each lone pair closes the bond angle by roughly 2 to 3°: $\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ about 107°, $\mathrm{H_2O}$ about 104.5°.
In the five-domain shape, lone pairs take the equatorial positions; in the six-domain shape, two lone pairs sit opposite each other.
$\mathrm{AlCl_3}$ on its own is trigonal planar; in $\mathrm{Al_2Cl_6}$ each Al has four Cl around it and is tetrahedral.

Domains	Lone pairs	Shape	Bond angle	Examples
2	0	Linear	180°	$\mathrm{BeCl_2}$ , $\mathrm{CO_2}$ , $\mathrm{C_2H_2}$
3	0	Trigonal planar	120°	$\mathrm{BF_3}$ , $\mathrm{SO_3}$ , $\mathrm{CO_3^{2-}}$
3	1	Bent	just under 120°	$\mathrm{SO_2}$ , $\mathrm{O_3}$
4	0	Tetrahedral	109.5°	$\mathrm{CH_4}$ , $\mathrm{NH_4^+}$ , $\mathrm{SO_4^{2-}}$
4	1	Trigonal pyramidal	about 107°	$\mathrm{NH_3}$ , $\mathrm{PCl_3}$ , $\mathrm{H_3O^+}$
4	2	Bent	about 104.5°	$\mathrm{H_2O}$ , $\mathrm{H_2S}$
5	0	Trigonal bipyramidal	120° and 90°	$\mathrm{PCl_5}$
5	1	Seesaw	below 120° and 90°	$\mathrm{SF_4}$
5	2	T-shaped	about 90°	$\mathrm{ClF_3}$
6	0	Octahedral	90°	$\mathrm{SF_6}$
6	2	Square planar	90°	$\mathrm{XeF_4}$

Practice this conceptself-check · 4 quick reps

Ammonia is pyramidal, not tetrahedral

Its four domains point to the corners of a tetrahedron, but one is a lone pair, and the shape is named from the atoms: trigonal pyramidal, about 107°. Four atoms around a centre with a lone pair is never trigonal planar either.

Concept 3 of 3: Polar and non-polar molecules: when bond dipoles cancel

Each polar bond is a small arrow pointing towards its more electronegative atom. If the arrows are arranged symmetrically, they pull equally in opposite directions and cancel, like a tug of war with equal teams. A lone pair or a mix of different outer atoms breaks the symmetry, and the molecule is left with a positive end and a negative end.

Definition

A molecule is polar (has a permanent dipole moment) when its bond dipoles do not cancel.

Step 1: are there any polar bonds? If none ( $\mathrm{Cl_2}$ , $\mathrm{CH_4}$ in practice), the molecule is non-polar.
Step 2: is the shape symmetric with identical outer atoms and no lone pair on the centre (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar)? Then the dipoles cancel: non-polar.
A lone pair on the central atom (bent, pyramidal) or different outer atoms around a symmetric centre usually leave a dipole: polar.
Polar bonds do not guarantee a polar molecule.

Molecule	Shape	Bond dipoles cancel?	Polar molecule?
$\mathrm{CO_2}$	Linear	Yes	No
$\mathrm{H_2O}$	Bent	No	Yes
$\mathrm{BF_3}$	Trigonal planar	Yes	No
$\mathrm{NH_3}$ , $\mathrm{PCl_3}$	Trigonal pyramidal	No	Yes
$\mathrm{CCl_4}$	Tetrahedral, four identical atoms	Yes	No
$\mathrm{CHCl_3}$ , $\mathrm{CH_3Cl}$	Tetrahedral, mixed atoms	No	Yes
$\mathrm{PCl_5}$ , $\mathrm{SF_6}$	Trigonal bipyramidal, octahedral	Yes	No
HCl, HBr	Linear, two different atoms	Only one bond	Yes

Practice this conceptself-check · 3 quick reps

Polar bonds can still give a non-polar molecule

The C to O bonds in

\mathrm{CO_2}

and the B to F bonds in

\mathrm{BF_3}

are strongly polar, but the shapes are symmetric and the dipoles cancel. Ask about the shape before answering "polar".

A tetrahedral molecule can be polar

Tetrahedral cancels only when all four outer atoms are the same.

\mathrm{CHCl_3}

is tetrahedral and polar, because its C to Cl and C to H bonds pull with different strengths.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Counting electron domains on the central atom (VSEPR)
Domains on the central atom
$D = \frac{V + m - q}{2}$

Reference tables (2)

The VSEPR shapes and their bond angles11 rows

Domains	Lone pairs	Shape	Bond angle	Examples
2	0	Linear	180°	$\mathrm{BeCl_2}$ , $\mathrm{CO_2}$ , $\mathrm{C_2H_2}$
3	0	Trigonal planar	120°	$\mathrm{BF_3}$ , $\mathrm{SO_3}$ , $\mathrm{CO_3^{2-}}$
3	1	Bent	just under 120°	$\mathrm{SO_2}$ , $\mathrm{O_3}$
4	0	Tetrahedral	109.5°	$\mathrm{CH_4}$ , $\mathrm{NH_4^+}$ , $\mathrm{SO_4^{2-}}$
4	1	Trigonal pyramidal	about 107°	$\mathrm{NH_3}$ , $\mathrm{PCl_3}$ , $\mathrm{H_3O^+}$
4	2	Bent	about 104.5°	$\mathrm{H_2O}$ , $\mathrm{H_2S}$
5	0	Trigonal bipyramidal	120° and 90°	$\mathrm{PCl_5}$
5	1	Seesaw	below 120° and 90°	$\mathrm{SF_4}$
5	2	T-shaped	about 90°	$\mathrm{ClF_3}$
6	0	Octahedral	90°	$\mathrm{SF_6}$
6	2	Square planar	90°	$\mathrm{XeF_4}$

Polar and non-polar molecules: when bond dipoles cancel8 rows

Molecule	Shape	Bond dipoles cancel?	Polar molecule?
$\mathrm{CO_2}$	Linear	Yes	No
$\mathrm{H_2O}$	Bent	No	Yes
$\mathrm{BF_3}$	Trigonal planar	Yes	No
$\mathrm{NH_3}$ , $\mathrm{PCl_3}$	Trigonal pyramidal	No	Yes
$\mathrm{CCl_4}$	Tetrahedral, four identical atoms	Yes	No
$\mathrm{CHCl_3}$ , $\mathrm{CH_3Cl}$	Tetrahedral, mixed atoms	No	Yes
$\mathrm{PCl_5}$ , $\mathrm{SF_6}$	Trigonal bipyramidal, octahedral	Yes	No
HCl, HBr	Linear, two different atoms	Only one bond	Yes

Watch out for (4)

Forgetting the charge changes the shape
→ Counting electron domains on the central atom (VSEPR)
Ammonia is pyramidal, not tetrahedral
→ The VSEPR shapes and their bond angles
Polar bonds can still give a non-polar molecule
→ Polar and non-polar molecules: when bond dipoles cancel
A tetrahedral molecule can be polar
→ Polar and non-polar molecules: when bond dipoles cancel

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